Equal valumes of 0.1 M urea and 0.1 M glucose are mixed. The mixture will have :

Options

A. lower osmotic pressure
B. same osmotic pressure
C. higher osmotic pressure
D. none of these

Answer

To solve the problem, we need to determine the osmotic pressure of the mixture formed by equal volumes of 0.1 M urea and 0.1 M glucose. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Components
We have two solutions being mixed:
- Urea solution: 0.1 M
- Glucose solution: 0.1 M

### Step 2: Identify the Formula for Osmotic Pressure
The osmotic pressure (\( \pi \)) of a solution can be calculated using the formula:
\[ \pi = C \cdot R \cdot T \cdot i \]
where:
- \( C \) = concentration of the solution
- \( R \) = gas constant
- \( T \) = temperature (in Kelvin)
- \( i \) = Van't Hoff factor (which indicates the number of particles the solute dissociates into)

### Step 3: Determine the Van't Hoff Factor
Both urea and glucose are non-electrolytes, meaning they do not dissociate into ions in solution. Therefore, the Van't Hoff factor (\( i \)) for both urea and glucose is:
\[ i = 1 \]

### Step 4: Calculate the Osmotic Pressure for Each Solution
Since both solutions have the same concentration (0.1 M), and the Van't Hoff factor is also the same (1), we can express the osmotic pressure for both solutions as follows:
- For urea:
\[ \pi_{\text{urea}} = 0.1 \cdot R \cdot T \cdot 1 = 0.1 \cdot R \cdot T \]
- For glucose:
\[ \pi_{\text{glucose}} = 0.1 \cdot R \cdot T \cdot 1 = 0.1 \cdot R \cdot T \]

### Step 5: Compare the Osmotic Pressures
Since both osmotic pressures are equal:
\[ \pi_{\text{urea}} = \pi_{\text{glucose}} \]

### Step 6: Conclusion
When equal volumes of 0.1 M urea and 0.1 M glucose are mixed, the resulting mixture will have the same osmotic pressure as each individual solution. Therefore, the answer is:
**The mixture will have the same osmotic pressure.**

---